[m190908m]

Is my thinking correct? Have I used the correct equations? Thank you
Question:
You wish to find the enthalpy of reaction per mole of nitric acid for the following reaction:

Mg (s) + 2HNO3 (aq) --> Mg(NO3)2 (aq) + H2 (g).

At 25 degrees C, you add 0.050 grams of magnesium flakes to a makeshift coffee-cup calorimeter (Ccal= 11 Joules/C) that contains 100.0 mL of 1.25M HNO3. You close the top and monitor the temperature of the solution until it stabilizes at 27.25 degrees C. Assume the density of the solution is 1.01 g/mL and that the specific heat is 4.22 J/(g*°C).

a) What constitutes the system in this experiment?

My answer: Everything INSIDE the coffee-cup calorimeter.

b) WHat constitutes the surroundings in this experiment?

My answer: Everything OUTside the coffee-cup calorimeter

c) How are the heat of reaction and the heat of the surroundings related?

My answer: Heat gained or lost from the system is proportional to the heat gained or lost from the surroundings.

d) Calculate the heat gained or lost by the calorimeter?

My work:

(11 J/°C)*(27.25°C-25°C) = 24.75 joules?

e) Calculate the heat gained or lost by the solution?

My work:

(101 grams soln)*(4.184 J/g*°C)*(27.25°C-25°C) = 950.8 joules?

f) Calculate the heat gained or lost by the reaction?

My thoughts: I'm assuming because my answers to question a, b, and c, that energy gained/lost from the calorimeter is the energy gained or lost by the reaction. Is my thinking correct?
...If so, then the heat gained or lost by the reaction is equal to -24.75 joules

g) Calculate the enthalpy of reaction per mole of nitric acid?

My work:

(0.050 g Mg (solid))*(1 mol Mg/24.3050 g Mg)*(2 mol HNO3/1 mol Mg)= .411 moles HNO3 -> .9508 kJ/.411 moles =  2.31 kJ/mol HNO3

So that's it, if your wondering what I'm asking/looking for is simply confirmation that my thinking is correct and I'm doing the problems correctly. Thank you in advance.

[Borek]

Heat gained or lost from the system is proportional to the heat gained or lost from the surroundings.

When you say "proportional" you suggest they are different. They are exactly the same.

My thoughts: I'm assuming because my answers to question a, b, and c, that energy gained/lost from the calorimeter is the energy gained or lost by the reaction. Is my thinking correct?
...If so, then the heat gained or lost by the reaction is equal to -24.75 joules

Where did the heat to heat up the solution came from?

[mjc123]

Quote from: m190908m on Today at 04:59:05 PM
Heat gained or lost from the system is proportional to the heat gained or lost from the surroundings.

When you say "proportional" you suggest they are different. They are exactly the same.

Actually not exactly the same, the proportionality constant is -1.
But you are not answering the question, which was not about the heat gain/loss of the system and surroundings, but about the heat of reaction and the heat of the surroundings.

Your answers to a) and b) - does the calorimeter itself count as part of the system or the surroundings?

[Borek]

Actually not exactly the same, the proportionality constant is -1.

This is tricky. I believe there are two conventions at work here, for me "heat lost" doesn't need sign change as the minus is already incorporated into the definition of a loss.

[tex]Q_{gained} = Q_{final} - Q_{initial}[/tex]

and

[tex]Q_{lost} = Q_{initial} - Q_{final}[/tex]

Thus

[tex]Q_{lost~by~the~system} = Q_{gained~by~the~surroundings}[/tex]

but

[tex]\Delta Q_{system} = - \Delta Q_{surroundings}[/tex]

I wonder if it is not something stemming from differences between Polish and English nomenclature.

[m190908m]

Quote from: m190908m on Today at 04:59:05 PM
Heat gained or lost from the system is proportional to the heat gained or lost from the surroundings.

When you say "proportional" you suggest they are different. They are exactly the same.

Actually not exactly the same, the proportionality constant is -1.
But you are not answering the question, which was not about the heat gain/loss of the system and surroundings, but about the heat of reaction and the heat of the surroundings.

Your answers to a) and b) - does the calorimeter itself count as part of the system or the surroundings?

I believe the calorimeter is part of the system, unless otherwise defined- like in this case. But I also believe perhaps the focus of this question, is the chemical reaction taking place in the calorimeter- the reaction between magnesium and nitric acid. So perhaps the calorimeter is in fact part of the surroundings? My only system is the chemical reaction.

[m190908m]

Actually not exactly the same, the proportionality constant is -1.

This is tricky. I believe there are two conventions at work here, for me "heat lost" doesn't need sign change as the minus is already incorporated into the definition of a loss.

[tex]Q_{gained} = Q_{final} - Q_{initial}[/tex]

and

[tex]Q_{lost} = Q_{initial} - Q_{final}[/tex]

Thus

[tex]Q_{lost~by~the~system} = Q_{gained~by~the~surroundings}[/tex]

but

[tex]\Delta Q_{system} = - \Delta Q_{surroundings}[/tex]

I wonder if it is not something stemming from differences between Polish and English nomenclature.

My thinking now is that because the focus is in fact on the chemical reaction taking place, the system. The calorimeter is an "isolated" container, but I'm not sure if I can call it my system, or even part of it? The heat from the reaction came from a chemical rxn between an acid and solid magnesium. I think its safe to say that the thermal energy lost from the chemical reaction is equal to the thermal energy "exiting" the calorimeter (spiking the thermometer?).

Also I know its my question, but I would agree with Borek. Mathematically it makes sense that Q_lost=Q_initial - Q_final, simply because a negative out front would flip the signs regardless. Q_lost=Q_initial - Q_final can also be written; -Q_lost=Q_{final - Qinitial}

[Borek]

I believe the calorimeter is part of the system, unless otherwise defined- like in this case. But I also believe perhaps the focus of this question, is the chemical reaction taking place in the calorimeter- the reaction between magnesium and nitric acid. So perhaps the calorimeter is in fact part of the surroundings? My only system is the chemical reaction.

Think about it this way: your system is everything that changes the temperature and is isolated from the surroundings. That means both the reaction mixture and the calorimeter, and they were both heated up by the reaction. Calorimeter is build to minimalize heat losses - you want all heat produced to be used to heat your system up, you don't want the heat to escape and heat up the surroundings. Unfortunately, you can't isolate the reaction mixture completely, at best, you need to heat up the container to some extent. That's where the heat capacity of the calorimeter comes in, but it only adds to the reaction mixture heat capacity.

[m190908m]

Thank you, that's actually a very good way to look at it. So just to put the nails in the coffin- I did use all the correct equations in the quantitative questions?

[Borek]

So just to put the nails in the coffin- I did use all the correct equations in the quantitative questions?

No, that's what we were pointing you to with earlier comments and questions.